Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $q = \dfrac{10a^2 + 30a}{a^2 + 12a + 27} \div \dfrac{-9a + 54}{2a + 18} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{10a^2 + 30a}{a^2 + 12a + 27} \times \dfrac{2a + 18}{-9a + 54} $ First factor the quadratic. $q = \dfrac{10a^2 + 30a}{(a + 9)(a + 3)} \times \dfrac{2a + 18}{-9a + 54} $ Then factor out any other terms. $q = \dfrac{10a(a + 3)}{(a + 9)(a + 3)} \times \dfrac{2(a + 9)}{-9(a - 6)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ 10a(a + 3) \times 2(a + 9) } { (a + 9)(a + 3) \times -9(a - 6) } $ $q = \dfrac{ 20a(a + 3)(a + 9)}{ -9(a + 9)(a + 3)(a - 6)} $ Notice that $(a + 3)$ and $(a + 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 20a(a + 3)\cancel{(a + 9)}}{ -9\cancel{(a + 9)}(a + 3)(a - 6)} $ We are dividing by $a + 9$ , so $a + 9 \neq 0$ Therefore, $a \neq -9$ $q = \dfrac{ 20a\cancel{(a + 3)}\cancel{(a + 9)}}{ -9\cancel{(a + 9)}\cancel{(a + 3)}(a - 6)} $ We are dividing by $a + 3$ , so $a + 3 \neq 0$ Therefore, $a \neq -3$ $q = \dfrac{20a}{-9(a - 6)} $ $q = \dfrac{-20a}{9(a - 6)} ; \space a \neq -9 ; \space a \neq -3 $